What is the coefficient of the term $x^3y^5$, as a result of the binomial expansion of the following term?

Question

We have the term $(1+xy+y^2)^n$ If we expand it using the binomial theorem, why is the factor of the term $x^3y^5$ the following: $4{n\choose 4}$? (The binomial coefficient n choose 4 multiplied by 4)

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffd,10px]{\bracks{x^{3}}\bracks{y^{5}}\pars{1 + xy + y^{2}}^{n}} \\[5mm] = &\ \bracks{x^{3}}\bracks{y^{5}}\sum_{a,b,c\ \in\ \mathbb{N}_{\geq\ 0}}{n! \over a!\, b!\, c!}\, 1^{a}\pars{xy}^{b}\pars{y^{5}}^{c}\bracks{a + b + c = n} \\[5mm] = &\ \bracks{x^{3}}\bracks{y^{5}}\sum_{a,b,c\ \in\ \mathbb{N}_{\geq\ 0}}{n! \over a!\, b!\, c!}\, x^{b}y^{b + 2c}\bracks{a + b + c = n} \\[5mm] = &\ \sum_{a,b,c\ \in\ \mathbb{N}_{\geq\ 0}}{n! \over a!\, b!\, c!}\, \bracks{b = 3}\ \underbrace{\bracks{b + 2c = 5}\bracks{a + b + c = n}} _{\ds{\implies \pars{~c = 1\ \mbox{and}\ a = n - 4}~}} \\[5mm] = &\ {n! \over \pars{n - 4}!\, 3!\, 1!} = {n! \over \pars{n - 4}!\, 4!}\,{4! \over 3!} = \bbx{4{n \choose 4}} \end{align}

Answer 2

Yes, for n = 5, use the binomial theorem by substituting $xy + y^2$ for "$a$" and then expand $(1 + a)^5 = 1 + 5a + 10a^2 + 10a^3 + 5a^4 + a^5$.

Then substitute back $xy + y^2$ for $a$.........

Here, the only term which will give $x^3y^5$ is $5a^4$ whereby again using the binomial theorem on $5(xy + y^2)^4$ we get......

$5x^4y^4 + 5(4)x^3y^5 + 5(6)x^2y^6 + 5(4)xy^7 + 5y^8$

We get $20x^3y^5$ which is in keeping with $4\binom{5}{4}$

So to answer you question in a general sense, expanding a trinomial involves more than one application of the binomial theorem for some of the terms.

Answer 3

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.

We obtain \begin{align*} \color{blue}{[x^3y^5]}&\color{blue}{(1+xy+y^2)}\\ &=[x^3y^5]\sum_{j=0}^n\binom{n}{j}(xy)^j(1+y^2)^{n-j}\tag{1}\\ &=[y^5]\binom{n}{3}y^3(1+y^2)^{n-3}\tag{2}\\ &=\binom{n}{3}[y^2]\sum_{j=0}^{n-3}\binom{n-3}{j}y^{2j}\tag{3}\\ &=\binom{n}{3}\binom{n-3}{1}\tag{4}\\ &=\frac{n(n-1)(n-2)}{3!}\cdot(n-3)\\ &=4\cdot\frac{n(n-1)(n-2)(n-3)}{4!}\\ &\,\,\color{blue}{=4\binom{n}{4}} \end{align*}

Comment:

• In (1) we apply the binomial theorem to $(xy+(1+y^2))^n$. This way we can easily select the coefficient of $x^3$.

• In (2) we select the coefficient of $x^3$.

• In (3) we use the formula $[y^{p-q}]A(y)=[y^p]y^qA(y)$ and apply the binomial theorem again.

• In (4) we select the coefficient of $y^2$.